Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Nov 1912:30 PM EST
-01:30 PM EST
Learn how Keshav, a Chartered Accountant, scored an impressive 705 on GMAT in just 30 days with GMATWhiz's expert guidance. In this video, he shares preparation tips and strategies that worked for him, including the mock, time management, and more
Nov 2001:30 PM EST
-02:30 PM IST
Learn how Kamakshi achieved a GMAT 675 with an impressive 96th %ile in Data Insights. Discover the unique methods and exam strategies that helped her excel in DI along with other sections for a balanced and high score.
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2407:00 PM PST
-08:00 PM PST
Full-length FE mock with insightful analytics, weakness diagnosis, and video explanations!
Nov 2510:00 AM EST
-11:00 AM EST
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
Updated on: 27 Feb 2019, 04:20
Originally posted by EgmatQuantExpert on 30 Nov 2018, 05:58.
Last edited by EgmatQuantExpert on 27 Feb 2019, 04:20, edited 1 time in total.
Last edited by EgmatQuantExpert on 27 Feb 2019, 04:20, edited 1 time in total.
Kudos
Bookmarks
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
33% (03:06) correct
67%
(02:46)
wrong
based on 203
sessions
History
Date
Time
Result
Not Attempted Yet
e-GMAT Question of the Week #25
For a meeting, 9 speakers must be selected to give a speech, one after the other, from a group of 12 people. What is the probability that A, B and C are three among the total 9 speakers selected and B speaks before A and C?
For a meeting, 9 speakers must be selected to give a speech, one after the other, from a group of 12 people. What is the probability that A, B and C are three among the total 9 speakers selected and B speaks before A and C?
- A. \(\frac{7}{12*11*10}\)
B. \(\frac{7}{12*11*5}\)
C. \(\frac{7}{11*3*5}\)
D. \(\frac{7}{11*10}\)
E. \(\frac{7}{11*5}\)
13 Jul 2024, 08:12
Kudos
Bookmarks
Let's break this into 2 parts: probability that A,B,C will be part of 9 people and probability that B speaks before A and C
Part 1: probability that A,B,C will be part of 9 people
Since A,B,C needs to be in 9 people the 3 spots are taken and rest 6 are up to grab in 12-3c6 i.e 9c6
Total number of ways in which 9 can be selected 12c9.
P(A) = 9c6/12c9 = 9*8*7/12*11*10
Part 2: We need to make sure B speaks before A and C.
Total number of arrangement possible with A,B,C are 3! = 6.
In that B should be front. B A/C C/A = 2!
P(B) = 2/6 = 1/3
now finally P(A)*P(B) = 9*8*7/12*11*10*3 = 7/11*5
Hence option E.
Part 1: probability that A,B,C will be part of 9 people
Since A,B,C needs to be in 9 people the 3 spots are taken and rest 6 are up to grab in 12-3c6 i.e 9c6
Total number of ways in which 9 can be selected 12c9.
P(A) = 9c6/12c9 = 9*8*7/12*11*10
Part 2: We need to make sure B speaks before A and C.
Total number of arrangement possible with A,B,C are 3! = 6.
In that B should be front. B A/C C/A = 2!
P(B) = 2/6 = 1/3
now finally P(A)*P(B) = 9*8*7/12*11*10*3 = 7/11*5
Hence option E.
eswarchethu135
GMAT 2: 640 Q49 V27
Posts: 276
06 Dec 2018, 00:03
Kudos
Bookmarks
For a meeting, 9 speakers must be selected to give a speech, one after the other, from a group of 12 people. What is the probability that A, B and C are three among the total 9 speakers selected and B speaks before A and C?
Firstly, 9 speakers are selected from 12. This can be done in 12*11*10 ways or \(^{12}P_9\) ways.
Now A,B and C are in the group of 9 people. and B should speak before A and C. This can be done in 7 ways.
A and C be seated after B in the following ways:
B_ _ _ _ _ _ _ _ = \(^8C_2*2!\) (Selecting 2 places among 8 and arranging A and C within their places)
_B_ _ _ _ _ _ _ = \(^7C_2*2!\)
_ _B_ _ _ _ _ _ = \(^6C_2*2!\)
_ _ _B_ _ _ _ _ = \(^5C_2*2!\)
_ _ _ _B_ _ _ _ = \(^4C_2*2!\)
_ _ _ _ _B_ _ _ = \(^3C_2*2!\)
_ _ _ _ _ _B_ _ = \(^2C_2*2!\)
Now B can't be in 8th place as it leaves only one place after B in which both A and C can't sit.
\(\frac{^8C_2*2! + ^7C_2*2! + ^6C_2*2! + ^5C_2*2! + ^4C_2*2! + ^3C_2*2! + ^2C_2*2!}{12*11*10}\)
\(\frac{(7*8)+(6*7)+(5*6)+(4*5)+(3*4)+(2*3)+(1*2)}{12*11*10}\)
\(\frac{56+42+30+20+12+6+2}{12*11*10}\)
\(\frac{168}{12*11*10}\)
\(\frac{7}{11*5}\)
Though it may seem lengthy, if you are confident about the procedure then it should not take more than 1 min 30 seconds to solve this.
OPTION: E
Firstly, 9 speakers are selected from 12. This can be done in 12*11*10 ways or \(^{12}P_9\) ways.
Now A,B and C are in the group of 9 people. and B should speak before A and C. This can be done in 7 ways.
A and C be seated after B in the following ways:
B_ _ _ _ _ _ _ _ = \(^8C_2*2!\) (Selecting 2 places among 8 and arranging A and C within their places)
_B_ _ _ _ _ _ _ = \(^7C_2*2!\)
_ _B_ _ _ _ _ _ = \(^6C_2*2!\)
_ _ _B_ _ _ _ _ = \(^5C_2*2!\)
_ _ _ _B_ _ _ _ = \(^4C_2*2!\)
_ _ _ _ _B_ _ _ = \(^3C_2*2!\)
_ _ _ _ _ _B_ _ = \(^2C_2*2!\)
Now B can't be in 8th place as it leaves only one place after B in which both A and C can't sit.
\(\frac{^8C_2*2! + ^7C_2*2! + ^6C_2*2! + ^5C_2*2! + ^4C_2*2! + ^3C_2*2! + ^2C_2*2!}{12*11*10}\)
\(\frac{(7*8)+(6*7)+(5*6)+(4*5)+(3*4)+(2*3)+(1*2)}{12*11*10}\)
\(\frac{56+42+30+20+12+6+2}{12*11*10}\)
\(\frac{168}{12*11*10}\)
\(\frac{7}{11*5}\)
Though it may seem lengthy, if you are confident about the procedure then it should not take more than 1 min 30 seconds to solve this.
OPTION: E
